Director Circle of an Ellipse

The general equation of a tangent line to the ellipse with a slope $$m$$ is:
$$y = mx \pm \sqrt{a^2m^2 + b^2}$$
If this tangent passes through an external intersection point $$T(h, k)$$, we substitute these coordinates into the line equation:
$$k - mh = \pm \sqrt{a^2m^2 + b^2}$$
$$(k - mh)^2 = a^2m^2 + b^2$$
$$k^2 - 2mhk + m^2h^2 = a^2m^2 + b^2$$
$$m^2(h^2 - a^2) - 2mhk + (k^2 - b^2) = 0$$
\Because the two tangents meeting at $$T$$ are perpendicular, the product of their slopes ($$m_1 \cdot m_2$$) must equal $$-1$$.($$\frac{C}{A}$$):
$$\frac{k^2 - b^2}{h^2 - a^2} = -1 \implies k^2 - b^2 = -h^2 + a^2 \implies h^2 + k^2 = a^2 + b^2$$
Generalizing $$(h,k)$$ to standard coordinates $$(x,y)$$ gives the equation of a circle: $$x^2 + y^2 = a^2 + b^2$$. 

Thus, the radius of this locus (the Director Circle) is $$\sqrt{a^2 + b^2}$$=5